Let ABCD be a trapezium with AD parallel to B | vedclass

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(B) Let $h$ be the height of the trapezium $ABCD$ (the perpendicular distance between $AD$ and $BC$).Let $AP \perp BC$ and $DQ \perp BC$. Thus,$AP = D

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$3$

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(B) Let $h$ be the height of the trapezium $ABCD$ (the perpendicular distance between $AD$ and $BC$).Let $AP \perp BC$ and $DQ \perp BC$. Thus,$AP = DQ = h$.Since $AB = AM$,$	riangle ABM$ is an isosceles triangle. In $	riangle ABM$,$AP$ is the altitude to the base $BM$,so $P$ is the midpoint of $BM$,i.e.,$BP = PM$.Area of $	riangle ABM = \frac{1}{2} 	imes BM 	imes h = \frac{1}{2} 	imes (2PM) 	imes h = PM 	imes h$.Similarly,since $DC = DM$,$	riangle DCM$ is an isosceles triangle. In $	riangle DCM$,$DQ$ is the altitude to the base $MC$,so $Q$ is the midpoint of $MC$,i.e.,$MQ = QC$.Area of $	riangle DCM = \frac{1}{2} 	imes MC 	imes h = \frac{1}{2} 	imes (2MQ) 	imes h = MQ 	imes h$.Area of $	riangle AMD = \frac{1}{2} 	imes AD 	imes h$.Since $AD$ is parallel to $BC$,$AD = PQ = PM + MQ$.Area of $	riangle AMD = \frac{1}{2} 	imes (PM + MQ) 	imes h = \frac{1}{2} 	imes PM 	imes h + \frac{1}{2} 	imes MQ 	imes h$.Area of trapezium $ABCD = 	ext{Area}(	riangle ABM) + 	ext{Area}(	riangle AMD) + 	ext{Area}(	riangle DCM) = PM 	imes h + \frac{1}{2}(PM + MQ)h + MQ 	imes h = \frac{3}{2}(PM + MQ)h$.Ratio $= \frac{	ext{Area}(ABCD)}{	ext{Area}(	riangle AMD)} = \frac{\frac{3}{2}(PM + MQ)h}{\frac{1}{2}(PM + MQ)h} = 3$.

Let $ABCD$ be a trapezium with $AD$ parallel to $BC$. Assume there is a point $M$ in the interior of the segment $BC$ such that $AB=AM$ and $DC=DM$. Then,the ratio of the area of the trapezium to the area of $\triangle AMD$ is

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